MA131 - Aalysis 1 Workbook 10 Series IV Autum 2004 Cotets 4.19 Rearragemets of Series...................... 1
4.19 Rearragemets of Series If you take ay fiite set of umbers ad rearrage their order, their sum remais the same. But the truly weird ad mid-bedig fact about ifiite sums is that, i some cases, you ca rearrage the terms to get a totally differet sum. We look at oe example i detail. The sequece (b ) = 1, 1 2, 1 4, 1 3, 1 6, 1 8, 1 5, 1 10, 1 12, 1 7, 1 14, 1 16, 1 9,... cotais all the umbers i the sequece (a ) = 1, 1 2, 1 3, 1 4, 1 5, 1 6, 1 7, 1 8, 1 9, 1 10, 1 11, 1 12, 1 13, 1 14,... but rearraged i a differet order: each of the positive terms is followed by ot oe but two of the egative terms. You ca also see that each umber i (b ) is cotaied i (a ). So this rearragemet effectively shuffles, or permutes, the idices of the origial sequece. This leads to the followig defiitio. Defiitio The sequece (b ) is a rearragemet of (a ) is there exists a bijectio σ : N N (i.e. a permutatio o N) such that b = a σ() for all. Assigmet 1 What permutatio σ has bee applied to the idices of the sequece (a ) to produce (b ) i the example above? Aswer this questio by writig dow a explicit formula for σ(3), σ(3 1), σ(3 2). Do t get hug up o this exercise if you re fidig it tricky, because the really iterestig part comes ext. We have defied the rearragemet of a sequece. Usig this defiitio, we say that the series b is a rearragemet of the series a if the sequece (b ) is a rearragemet of the sequece (a ). We kow already that: ( 1) +1 = 1 1 2 + 1 3 1 4 + 1 5 = log(2) We ow show that our rearragemet of this series has a differet sum. Shufflig the (Ifiite) Pack The permutatio σ simply shuffles about the terms of the old sequece (a ) to give the ew sequece (a σ() ). Reciprocal Rearragemets If (b ) is a rearragemet of (a ) the (a ) must be a rearragemet of (b ). Specifically, if b = a σ() the a = b σ 1 (). 1
Assigmet 2 Show that: b = 1 + 1 2 + 1 4 + 1 3 + 1 6 + 1 8 + 1 5 + 1 10 + 1 12 + 1 7 + 1 14 + 1 16 + 1 9 + = log 2 2 Hit: Let s = k=1 a k ad let t = k=1 b k. Show that t 3 = s2 2 by usig the followig groupig of the series b : ( 1 1 ) 1 ( 1 2 4 + 3 1 ) 1 ( 1 6 8 + 5 1 ) 1 ( 1 10 12 + 7 1 )... 14 This example is rather scary. However, for series with all positive terms it does ot matter i what order you add the terms. Lemma Suppose a is a coverget series of o-egative terms. If (b ) is a rearragemet of (a ) the b coverges ad b = a. Assigmet 3 Prove the lemma usig the followig steps for s = r=1 a r, t = r=1 b r ad A = a : 1. Suppose that the first N terms of the sequece of b s is icluded withi the first M terms of the sequece of a s. You ca defie M = max{σ(r) : r N}. Prove that t N s M A. Deduce that b is coverget to a sum B say, ad that B A. 2. Now reverse the above argumet with the first N terms of the a s icluded withi the first L terms of the b s to deduce that A B. Nor does it matter what order you add the terms of a absolutely coverget series. Theorem Suppose a is a absolutely coverget series. If (b ) is a rearragemet of (a ) the b is coverget ad b = a. 2
Assigmet 4 Prove the theorem. The followig steps will help. Let a be a absolutely coverge series ad b a rearragemet of the same series. Let u = 1 2 ( a + a ) v = 1 2 ( a a ), x = 1 2 ( b + b ) y = 1 2 ( b b ). 1. Why are the two series u ad v ecessarily coverget? 2. Are they both sequece of positive terms? 3. How does u relate to x ad v to y? 4. Prove that a = (u v ) = (s y ) = b. I 1837 the mathematicia Dirichlet discovered which type of series could be rearraged to give a differet total ad the result was displayed i a startlig form i 1854 by Riema. To describe their results we have oe fial defiitio. Defiitio The series a is said to be coditioally coverget if a is coverget but a is ot. Example Back to our familiar example: ( 1) +1 is coditioally coverget, because ( 1) +1 is coverget, but ( 1)+1 = 1 is ot. Exercise 1 Check from the defiitios that every coverget series is either absolutely coverget or is coditioally coverget. Assigmet 5 List which of the followig series are coditioally coverget. 1. ( 1) +1 2 2. cos(π) 3. ( 1) +1 1 + 2 Coditioally coverget series are the hardest to deal with ad ca behave very stragely. The key to uderstadig them is the followig lemma. Lemma If a series is coditioally coverget, the the series formed from just its positive terms teds to ifiity ad the series formed from just its egative terms teds to mius ifiity. Assigmet 6 Do Bur C5Q71 (Old Ed. C5Q67) ad prove this result. 3
Theorem Riema s Rearragemet Theorem Suppose a is a coditioally coverget series. The for every real umber l there is a rearragemet (b ) of (a ) such that b = l. The last lemma allows us to costruct a proof of the theorem alog the followig lies: We sum eough positive values to get us just above l. The we add eough egative values to take us back dow just below l. The we add eough positive terms to get back just above l agai, ad the eough egative terms to get back dow just below l. We repeat this idefiitely, i the process producig a rearragemet of a which coverges to l. Proof. Let (p ) be the subsequece of (a ) cotaiig all its positive terms, ad let (q ) be the subsequece of egative terms. First suppose that l 0. Sice p teds to ifiity, there exists N such that N i=1 p i > l. Let N 1 be the smallest such N ad let S 1 = N 1 i=1 p i. The S 1 = N 1 i=1 p i > l ad N1 1 i=1 p i l. Thus S 1 = N 1 1 i=1 p i +p N1 l +p N1, therefore 0 S 1 l p N1. To the sum S 1 we ow add just eough egative terms to obtai a ew sum T 1 which is less tha l. I other words, we choose the smallest iteger M 1 for which T 1 = S 1 + M 1 i=1 q i < l. This time we fid that 0 l T 1 q M1. We cotiue this process idefiitely, obtaiig sums alterately smaller ad larger tha l, each time choosig the smallest N i or M i possible. The sequece: The Ifiite Case We ca also rearrage ay coditioally coverget series to produce a series that teds to ifiity or mius ifiity. How would you modify the proof to show this? p 1,..., p N1, q 1,... q M1, p N1+1,... p N2, q M1+1,..., q M2,... is a rearragemet of (a ). Its partial sums icrease to S 1, the decrease to T 1, the icrease to S 2, the decrease to T 2, ad so o. To complete the proof we ote that for all i, S i l p Ni ad T i l q Mi. Sice a is coverget, we kow that (a ) is ull. It follows that subsequeces (p Ni ) ad (q Mi ) also ted to zero. This i tur esures that the partial sums of the rearragemet coverge to l, as required. I the case l < 0 the proof looks almost idetical, except we start off by summig eough egative terms to get us just below l. Assigmet 7 Draw a diagram which illustrates this proof. Make sure you iclude the limit l ad some poit S 1, T 1, S 2, T 2,.... All Wrapped Up Each coverget series is either coditioally coverget or absolutely coverget. Give the defiitio of these terms, there are o other possibilities. This theorem makes it clear that coditioally coverget series are the oly coverget series whose sum ca be perturbed by rearragemet. 4
Check Your Progress By the ed of this Workbook you should be able to: Defie what is meat by the rearragemet of a sequece or a series. Give a example of a rearragemet of the series ( 1) +1 sums to a differet value. = log 2 which Prove that if a is a series with positive terms, ad (b ) is a rearragemet of (a ) the b = a. Prove that if a is a absolutely coverget series, ad (b ) is a rearragemet of (a ) the b = a. Coclude that coditioally coverget series are the oly coverget series whose sum ca be altered by rearragemet. Kow that if a is a coditioally coverget series, the for every real umber l there is a rearragemet (a ) of (a ) such that b = l. 5